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3 years ago

310/5=?/1 find the unknown quantity

Mathematics

2 answers:

xeze [42]3 years ago

7 0

Answer:

the answer is 62

Step-by-step explanation:

310 divided by 5 = 62

62 divided by 1 = 62

Travka [436]3 years ago

7 0

Answer:

Step-by-step explanation:

Since the question is a like-fraction, 5 divided by 5 is 1, and 310 divided by 5 is 62, giving you your answer.

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Find the common difference of the arithmetic sequence -2,6,14,...

yawa3891 [41]

Lets note the change between each of the numbers in the sequence:

-2 --> 6 ---> 14

Next, make an equation between each of these numbers with x being the change between them

-2+ x = 6, 6 + x = 14 .

Now, solve each of these equations:

-2+ x = 6 --> -2+ x = 6 Add 2 on both sides to cancel out the -2 --> x = 8

6+ x = 14 --> -2+ x = 6 Minus 6 on both sides to cancel out the 6 --> x = 8

From this, the common difference will be 8.

Hope that helps.

8 0

3 years ago

Represent this statement as an equation: The difference of twice a number and 3 is 11.

yan [13]

Answer:

2x - 3 = 11

Step-by-step explanation:

"difference" means subtraction, so it is subtraction.

"twice a number" simply means 2x, since you do not know what the specified number is, it is x. and the 2 and just making it twice the number.

"Is 11" just means equals 11.

so, 2x - 3 = 11.

hope this helps!

3 0

4 years ago

Step 3: Let LM = x. We know the lengths of the radii of each circle, so KL = 12 +

kow [346]

Answer:

Step-by-step explanation:

Step 3:

Let LM = x

OK = KP = 12 units [Radii of circle K]

LN = LP = 8 units [Radii of circle L]

Therefore, KL = KP + PL

KL = 12 + 8

= 20 units

Step 4:

Since, ΔKOM and ΔLNM are the similar triangles,

By the property of two similar triangles, corresponding sides of these similar triangles will be proportional.

$\frac{OK}{NL}=\frac{KM}{LM}$

$\frac{12}{8}=\frac{x+20}{x}$

12x = 8(x + 20) [By cross multiplication]

12x = 8x + 160

12x - 8x = 160

4x = 160

x = 40

3 0

3 years ago

Find the measure of each side indicated. Round to the nearest tenth.

choli [55]

Answer:

I'm say d last choice is yours answer

7 0

4 years ago

Consider the accompanying data on breaking load (kg/25mm width) for various fabrics in both an unabraded condition and an abrade

WITCHER [35]

Answer:

The critical value for a one-tailed test with 7 degrees of freedom and level of significance α=0.01 is tc=2.998.

The test statistic (t=1.729) is below the critical value and falls within the acceptance region, so it is failed to reject the null hypothesis.

At a significance level of 0.01, there is not enough evidence to support the claim that the true difference is significantly bigger than 0.

Step-by-step explanation:

We have a matched-pair t-test for the difference.

We have the null and alternative hypothesis written as:

$H_0: \mu_d=0\\\\H_a: \mu_d>0$

We have n=8 pairs of data. We calculate the difference as:

$d_1=U_1-A_1=36.4-28.5=7.9$

Then, with this procedure we get the sample for d:

$d=[7.9,\, 35,\, 5.5,\, 4.2,\, 6.7,\, -3.7,\, -0.9,\, 3.3]$

The sample mean and standard deviation are:

$M=\dfrac{1}{n}\sum_{i=1}^n\,x_i\\\\\\M=\dfrac{1}{8}(7.9+35+5.5+. . .+3.3)\\\\\\M=\dfrac{58}{8}\\\\\\M=7.25\\\\\\s=\sqrt{\dfrac{1}{n-1}\sum_{i=1}^n\,(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{7}((7.9-7.25)^2+(35-7.25)^2+(5.5-7.25)^2+. . . +(3.3-7.25)^2)}\\\\\\s=\sqrt{\dfrac{985.08}{7}}\\\\\\s=\sqrt{140.73}=11.86\\\\\\$

Now, we can perform the one-tailed hypothesis test.

The significance level is 0.01.

The sample has a size n=8.

The sample mean is M=7.25.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=11.86.

The estimated standard error of the mean is computed using the formula:

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{11.86}{\sqrt{8}}=4.1931$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{7.25-0}{4.1931}=\dfrac{7.25}{4.1931}=1.729$

The degrees of freedom for this sample size are:

$df=n-1=8-1=7$

This test is a right-tailed test, with 7 degrees of freedom and t=1.729, so the P-value for this test is calculated as (using a t-table):

$\text{P-value}=P(t>1.729)=0.0637$

As the P-value (0.0637) is bigger than the significance level (0.01), the effect is not significant.

The null hypothesis failed to be rejected.

At a significance level of 0.01, there is not enough evidence to support the claim that the true difference is significantly bigger than 0.

If we use the critical value approach, the critical value for a one-tailed test with 7 degrees of freedom and level of significance α=0.01 is tc=2.998. The test statistic is below the critical value and falls within the acceptance region, so it is failed to reject the null hypothesis.

8 0

3 years ago